Download `A Simulation of Natural Selection` Lab Activity

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A Simulation of Natural Selection
Introduction
Evolution is a process that changes the genetic makeup of a population over time . Presumably, those genetic
changes are reflected in changes in the phenotypic makeup (the observable characteristics) of the population .
This exercise will demonstrate the effect of natural selection on the frequencies of three populations of
"beetles" . Natural selection, as formulated by Charles Darwin in Origin of Species (1859), is the most important
cause of evolution .
An individual's ability to reproduce depends on its ability to survive . If all gene variations conferred on every
individual the same capability to survive and reproduce, then the composition of a population would never
change . If a variation of a characteristic increases an individual's ability to survive or allows it to have more
offspring, that variation will be naturally selected . Darwin reasoned that the environment controlled in nature
what plant and animal breeders controlled artificially . Given an overproduction of offspring, natural variation
within a species and limited resources, the environment would "select" for those individuals whose traits would
enable a higher chance of survival and hence more offspring in the next generation .
Purpose
In this activity we will use three different beans to represent heritable variations in beetle morphology (size and
coloration of carapace) . These three beetle morphs will be studies in two different habitats .
Outline of Procedure
1 . Work with 1 or 2 partners . Count out exactly 10 each of lima, pinto and kidney beans . These represent each
beetle type ; the three types are equally common initially .
2 . You will choose 2 habitats and perform the procedure the same way in each one . Choose 2 from this list :
sidewalk, asphalt, brick, sand (or dirt) . The point is that the two substrates must be of different colors . Label
your data table with the type of habitat for both Habitat #1 and Habitat #2 .
3 . Scatter your beans randomly over an area about one square meter . This will be your predator foraging area .
The beans must be scattered (tossed), not dumped in a small pile .
4 . One person will be the designated predator for the habitat . (Switch roles for the second habitat) . The
predator will "eat" (pick up) exactly 20 beans . Remember to think like a predator, i .e. pick up what you see first as
quickly as you can . Place the 20 beans picked up in the plastic bag provided . Leave the rest of the beans on
the ground . They have survived the predator and will reproduce (their offspring will be represented by adding
beans to adjust the population size back up to 30, line F) .
5. Count the number of each bean collected (make sure you have exactly 20) and record the numbers on line B .
6 . Subtract the number of each kind eaten (line B) from the number you started with (line A), to obtain the
number of survivors (line C) .
7 . Assume that each survivor has two offspring . Record those values in line D . These are the numbers of each
bean that need to be scattered with the survivors to bring the population back up to exactly 30 . Count out and
scatter the required number of beans into the same area as your P1 survivors . Now complete line A by adding
lines C and D . These are your P2 or second generation populations .
8 . Repeat steps 4 through 7 two more times and complete the table for Habitat #1 . Remember, the offspring
values tell you how many beans of each type need to be scattered into your predator foraging area.
Pick up all your beans when you are finished . Repeat the entire procedure in Habitat #2 .
ResultsandAnalysis
We
will determine
whether evolution has occurred by' comparing the frequencies of each bean morph as the
beginning and at the end of our predation experiment . We will use a Chi Square Statistic
whether or not our final frequencies are significantly different from our initial ones .
(x2)
to determine
o = observed count in a category
e = expected count in that category
E means to sum for each category
Chi square = E (o-e)2
e
The Chi square value that you get is located on the Chi square table below . The degrees of freedom is one less
than the number of phenotypic categories . We have 3 phenotypic categories (the 3 bean morphs) .
Degrees of Freedom
1
2
3
4
$
6
7
8
9
10
15
20
25
30
P = .99
.00157
.020
.115
.297
.554
.672
1 .239
1 .646
2 .088
2 .556
5 .229
6 .260
11 .524
14 .953
.95
.00393
.103
.352
.711
1 .145
1 .635
2.167
2.733
3 .325
3 .040
7 .261
10 .851
14 .611
18.493
.60
.50
.20
.05
.01
.0642
.446
1 .005
1 .649
2 .343
3.070
3.122
4.594
5.310
6 .179
10 .307
14 .578
11 .940
23 .364
.455
1 .366
2 .366
3 .357
4 .351
5.348
6.346
7.344
8.343
9 .342
14 .339
19 .337
24 .337
29 .336
1 .642
3219
4.642
5 .989
7 .289
1 .551
9 .803
11 .030
12 .242
13.442
19.311
25.038
30.675
36.250
3 .641
5.991
7.115
9.488
11 .070
12 .592
14 .067
15 .507
16 .119
18 .307
24 .996
31 .410
37.652
43.773
6 .635
9 .210
11 .345
13 .277
15 .086
16 .812
16 .475
20 .090
21 .666
23 .209
30.578
37.566
44 .314
50 .892
A probability of less than .05 tells you that there is less than a 5% chance that the differences between our
beginning and ending counts could be due to random variation . This means that there is a 95% probability that
the differences are not due to random factors but are the result of our experiment .
Any statistic can only test two hypotheses, the null hypothesis of no difference and the alternative
hypothesis of significant difference . These statistical hypotheses can be summarized below :
H o (the null hypothesis) : there is no difference between our beginning and ending counts .
H1 (the alternative hypothesis) : there is a statistically significant difference between our beginning and
ending counts .
-•• Note that in science we can never prove that any hypothesis is true since there are always more data to
gather ; we can only prove hypotheses to be false .
A statistical hypothesis is not the same as an experimental hypothesis ; our experimental hypothesis involves
the role of predation and habitat in natural selection . What hypotheses were tested in this simulation?
Discussion and Conclusion
How do your results differ for the two habitats? How can you explain your results in terms of the hypotheses?
Short Individual Lab Write Urn On a single sheet of paper (word processed or printed in black ink) write :
1 . Title of lab and your name, date and APES period .
2 . A paragraph describing the specific hypotheses tested in the simulation .
3 . A paragraph describing the meaning of the results of your Chi square statistics .
4 . A paragraph answering the discussion questions .
5 . A well crafted conclusion of no more than two sentences .
You should label each written section, i.e . Title, Hypothesis, Results, Discussion, Conclusion .
Staple your data/chi square sheet to the back of your paper .
"This write up will be due two days after we finish the simulation .
Name :
-Group
members:
Date:
Summary of Beetle Captures
Habitat #1
Lima
Pinto
10
10
P.1
Habitat #2
Kidney
-Lima-
Pinto
Kidney Total
10
30
20
"eaten"
C#)
Total
1
"survivors" (A - 4)
10
"offspring" (2C)
20
P2 (C +xD)
30
30
"eaten"
20 ,
20
"survivors" (f,- F)
10
10
"offspring" (2G)
'20
20 .
P3 (G +H)
30
30
"eaten"
20
20
20
20
30
30
"survivors" (1 -:J)
I
L
"offspring" (2K)
I
P4 (K + L)
C
J
?(o
I'
L1)
err.Sfin -
Texfbvo/<_5, .4 Kc
Su (('jr, 1
1lJG
~
~
.
I
~~
X 2 ANALYSIS OF BEETLE CAPTURES
HABITAT
PHENOTYPE
EXPECTED (F 1 )
OBSERVED (F 4 )
EXPECTED (F 1 )
OBSERVED (F 4 )
X2
kidney
pinto
lima
TOTAL
HABITAT .
PHENOTYPE
kidney
pinto
lima
TOTAL
f